Replace $A$, $B$, $C$ by $0$, $1$, $2$. Then $y_i \equiv -x_i - x_{i+1} \pmod{3}$. We first show that such an $m$ does not exist for even $n$ as seen from the following $n$-tuple
$$(1, 0, 1, 0, \dots, 1, 0).$$
Next we show that $m$ exists when $n$ is odd. Since the total number of tuples that can be formed is $3^n$, there exist indices $s < t$ such that $R_s = R_t$. Without loss of generality, we may assume that $R_s$ is the only term in the sequence from $R_0$ to $R_t$ that is the same as $R_t$. If we can show that for any $i \ge 0$, $R_i$ can be obtained from $R_{i+1}$, then the sequence going backwards from $R_s$ is the same as the sequence going backwards from $R_t$. Since going backwards from $R_s$, we can get $R_0$, then going backwards from $R_t$, before reaching $R_s$, we can get an $n$-tuple which is the same as $R_0$. In other words, there is an index $m$, $s < m < t$, such that $R_0 = R_m$ and we are done.
To show that the sequence can be reversed, all we need to show is if two $n$-tuples $(a_1, a_2, \dots, a_n)$ and $(b_1, b_2, \dots, b_n)$ give rise to the same $n$-tuple, then they are in fact the same. We have from the given rule:
$$\text{-a 1 - a 2 = -b 1 - b 2 a 2 + a 3 = b 2 + b 3 -a n-2 - a n-1 = -b n-2 - b n-1 a n-1 + a n = b n-1 + b n -a n - a 1 = -b n - b 1}$$
Summing up, we get $2a_1 = 2b_1$ and therefore $a_1 = b_1$. Repeat this starting with the index pair $(2, 3)$, we can show that $a_2 = b_2$. Continuing this way, we have $(a_1, \dots, a_n) = (b_1, \dots, b_n)$: